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Alkenes. Addition Reactions Synthesis. Bonding in Alkenes. Electrons in pi bond are loosely held. The double bond acts as a nucleophile attacking electrophilic species. Carbocations are intermediates in the reactions. These reactions are called electrophilic additions.
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Alkenes Addition Reactions Synthesis
Bonding in Alkenes • Electrons in pi bond are loosely held. • The double bond acts as a nucleophile attacking electrophilic species. • Carbocations are intermediates in the reactions. • These reactions are called electrophilic additions.
Electrophilic Addition • Step 1: Pi electrons attack the electrophile. • Step 2: Nucleophile attacks the carbocation.
Addition of HX to Alkenes • Step 1 is the protonation of the double bond. • The protonation step forms the most stable carbocation possible. • In step 2, the nucleophile attacks the carbocation, forming an alkyl halide. • HBr, HCl, and HI can be added through this reaction.
Mechanism of Addition of HX Step 1: Protonation of the double bond. Step 2: Nucleophilic attack of the halide on the carbocation.
Regioselectivity • Markovnikov’s Rule: The addition of a proton to the double bond of an alkene results in a product with the acidic proton bonded to the carbon atom that already holds the greater number of hydrogens. • Markovnikov’s Rule (extended): In an electrophilic addition to the alkene, the electrophile adds in such a way that it generates the most stable intermediate.
Markovnikov’s Rule The acid proton will bond to carbon 3 in order to produce the most stable carbocation possible.
Free-Radical Addition of HBr • In the presence of peroxides, HBr adds to an alkene to form the “anti-Markovnikov” product. • Peroxides produce free radicals. • Only HBr has the right bond energy. • The HCl bond is too strong, so it will add according to Markovnikov’s rule, even in the presence of peroxide. • The HI bond tends to break heterolytically to form ions, it too will add according to Markovnikov’s rule.
Free-Radical Initiation • The peroxide bond breaks homolytically to form the first radical: • Hydrogen is abstracted from HBr.
Propagation Steps • Bromine adds to the double bond forming the most stable radical possible: • Hydrogen is abstracted from HBr:
Anti-Markovnikov Stereochemistry • The intermediate tertiary radical forms faster because it is more stable.
Addition of Halogens • Cl2, Br2, and sometimes I2 add to a double bond to form a vicinal dibromide. • This is an anti addition of halides.
Mechanism of Halogen Addition to Alkenes • The intermediate is a three-membered ring called the halonium ion.
Test for Unsaturation • Add Br2 in CCl4 (dark, red-brown color) to an alkene. • The color quickly disappears as the bromine adds to the double bond (left-side test tube). • If there is no double bond present the brown color will remain (right side). • “Decolorizing bromine” is the chemical test for the presence of a double bond.
Formation of Halohydrin • If a halogen is added in the presence of water, a halohydrin is formed. • Water is the nucleophile. • This is a Markovnikov addition: The bromide (electrophile) will add to the less substituted carbon.
Solved Problem 6 Propose a mechanism for the reaction of 1-methylcyclopentene with bromine water. Solution 1-Methylcyclopentene reacts with bromine to give a bromonium ion. Attack by water could occur at either the secondary carbon or the tertiary carbon of the bromonium ion. Attack actually occurs at the more substituted carbon, which bears more of the positive charge. The product is formed as a racemic mixture.
Solved Problem 7 When cyclohexene is treated with bromine in saturated aqueous sodium chloride, a mixture of trans-2-bromocyclohexanol and trans-1-bromo-2-chlorocyclohexane results. Propose a mechanism to account for these two products. Solution Cyclohexene reacts with bromine to give a bromonium ion, which will react with any available nucleophile. The most abundant nucleophiles in saturated aqueous sodium chloride solution are water and chloride ions. Attack by water gives the bromohydrin, and attack by chloride gives the dihalide. Either of these attacks gives anti stereochemistry.
Definitions • Stereospecific – only one stereoisomer is formed at the expense of the other (e.g. trans vs. cis) • Stereoselective – one stereoisomer is formed preferentially over the other.
Bromohydrin FormationAddition of Br – OHStereospecific & Regiospecific
Unsymmetrical Bromonium ionH2O opens ring at more hindered site
Hydration of Alkenes • The Markovnikov addition of water to the double bond forms an alcohol. • This is the reverse of the dehydration of alcohol. • Uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium toward hydration.
Orientation of Hydration The protonation follows Markovnikov’s rule: The hydrogen is added to the less substituted carbon in order to form the most stable carbocation. ,
Rearrangements Rearrangement: • Rearrangements can occur when there are carbocation intermediates. • A methyl shift after protonation will produce the more stable tertiary carbocation.
Solved Problem 1 Show how you would accomplish the following synthetic conversions. (a) Convert 1-methylcyclohexene to 1-bromo-1-methylcyclohexane. Solution This synthesis requires the addition of HBr to an alkene with Markovnikov orientation. Ionic addition of HBr gives the correct product.
Solved Problem 2 Convert 1-methylcyclohexanol to 1-bromo-2-methylcyclohexane. Solution This synthesis requires the conversion of an alcohol to an alkyl bromide with the bromine atom at the neighboring carbon atom. This is the anti-Markovnikov product, which could be formed by the radical-catalyzed addition of HBr to 1-methylcyclohexene. 1-Methylcyclohexene is easily synthesized by the dehydration of 1-methylcyclohexanol. The most substituted alkene is the desired product.
Solved Problem 2 (Continued) Solution (Continued) The two-step synthesis is summarized as follows:
